ratiotest and radius of convergence of Taylor expansion of $xsin(x)$
Today I helped a student who did not understand Taylor approximations. One of the exercises he had trouble with, was to determine the Taylor series of the function $$f: mathbb{R} to mathbb{R}: x mapsto xsin(x)$$
around $0$.
The Taylor-approximation only has even powers of $x$. The solution he had, determined the radius of convergence using the ratio test. However, the ratio test uses consecutive coefficients $c_k$ (if the power series is $sum c_k(x-a)^k$ and all $c_{2k+1}$ are zero. In the solution, the ratio of $c_{2k}/c_{2k+2}$ is computed and its limit is taken.
My question: is there a theorem which states that we can 'skip' coefficients because they are zero?
taylor-expansion
asked Dec 12 '18 at 22:36
StudentStudent
2,1121627
add a comment |
Today I helped a student who did not understand Taylor approximations. One of the exercises he had trouble with, was to determine the Taylor series of the function $$f: mathbb{R} to mathbb{R}: x mapsto xsin(x)$$
around $0$.
The Taylor-approximation only has even powers of $x$. The solution he had, determined the radius of convergence using the ratio test. However, the ratio test uses consecutive coefficients $c_k$ (if the power series is $sum c_k(x-a)^k$ and all $c_{2k+1}$ are zero. In the solution, the ratio of $c_{2k}/c_{2k+2}$ is computed and its limit is taken.
My question: is there a theorem which states that we can 'skip' coefficients because they are zero?
taylor-expansion
asked Dec 12 '18 at 22:36
StudentStudent
2,1121627
add a comment |
Today I helped a student who did not understand Taylor approximations. One of the exercises he had trouble with, was to determine the Taylor series of the function $$f: mathbb{R} to mathbb{R}: x mapsto xsin(x)$$
around $0$.
The Taylor-approximation only has even powers of $x$. The solution he had, determined the radius of convergence using the ratio test. However, the ratio test uses consecutive coefficients $c_k$ (if the power series is $sum c_k(x-a)^k$ and all $c_{2k+1}$ are zero. In the solution, the ratio of $c_{2k}/c_{2k+2}$ is computed and its limit is taken.
My question: is there a theorem which states that we can 'skip' coefficients because they are zero?
taylor-expansion
asked Dec 12 '18 at 22:36
StudentStudent
2,1121627
Today I helped a student who did not understand Taylor approximations. One of the exercises he had trouble with, was to determine the Taylor series of the function $$f: mathbb{R} to mathbb{R}: x mapsto xsin(x)$$
around $0$.
The Taylor-approximation only has even powers of $x$. The solution he had, determined the radius of convergence using the ratio test. However, the ratio test uses consecutive coefficients $c_k$ (if the power series is $sum c_k(x-a)^k$ and all $c_{2k+1}$ are zero. In the solution, the ratio of $c_{2k}/c_{2k+2}$ is computed and its limit is taken.
My question: is there a theorem which states that we can 'skip' coefficients because they are zero?
taylor-expansion
taylor-expansion
asked Dec 12 '18 at 22:36
StudentStudent
2,1121627
asked Dec 12 '18 at 22:36
StudentStudent
2,1121627
asked Dec 12 '18 at 22:36
StudentStudent
2,1121627
asked Dec 12 '18 at 22:36
StudentStudent
2,1121627
asked Dec 12 '18 at 22:36
StudentStudent
2,1121627
2,1121627
add a comment |
add a comment |
1 Answer
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votes
There is the root test.
Here the best thing is to note that the whole series can be written as a power series in $x^2$, and using the ratio test on this new power series yields the result for the new power series, thus for the original one.
answered Dec 12 '18 at 23:11
MindlackMindlack
2,33217
Thanks! One more question: suppose that the radius of convergence would turn out to be finite using the root-test and after rewritting as powers of $x^2$, do I have to take this into account (i.e. does the radius of convergence needs to be squared or something in that case?)
– Student
Dec 12 '18 at 23:27
Yes: if the radius of $sum_n{a_nx^n}$ is $R geq 0$, the radius of $sum_n{a_nx^{2n}}$ is $R^{1/2}$.
– Mindlack
Dec 12 '18 at 23:29
Thanks a lot!..
– Student
Dec 12 '18 at 23:29
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There is the root test.
Here the best thing is to note that the whole series can be written as a power series in $x^2$, and using the ratio test on this new power series yields the result for the new power series, thus for the original one.
answered Dec 12 '18 at 23:11
MindlackMindlack
2,33217
Thanks! One more question: suppose that the radius of convergence would turn out to be finite using the root-test and after rewritting as powers of $x^2$, do I have to take this into account (i.e. does the radius of convergence needs to be squared or something in that case?)
– Student
Dec 12 '18 at 23:27
Yes: if the radius of $sum_n{a_nx^n}$ is $R geq 0$, the radius of $sum_n{a_nx^{2n}}$ is $R^{1/2}$.
– Mindlack
Dec 12 '18 at 23:29
Thanks a lot!..
– Student
Dec 12 '18 at 23:29
add a comment |
There is the root test.
Here the best thing is to note that the whole series can be written as a power series in $x^2$, and using the ratio test on this new power series yields the result for the new power series, thus for the original one.
answered Dec 12 '18 at 23:11
MindlackMindlack
2,33217
Thanks! One more question: suppose that the radius of convergence would turn out to be finite using the root-test and after rewritting as powers of $x^2$, do I have to take this into account (i.e. does the radius of convergence needs to be squared or something in that case?)
– Student
Dec 12 '18 at 23:27
Yes: if the radius of $sum_n{a_nx^n}$ is $R geq 0$, the radius of $sum_n{a_nx^{2n}}$ is $R^{1/2}$.
– Mindlack
Dec 12 '18 at 23:29
Thanks a lot!..
– Student
Dec 12 '18 at 23:29
add a comment |
There is the root test.
Here the best thing is to note that the whole series can be written as a power series in $x^2$, and using the ratio test on this new power series yields the result for the new power series, thus for the original one.
answered Dec 12 '18 at 23:11
MindlackMindlack
2,33217
There is the root test.
Here the best thing is to note that the whole series can be written as a power series in $x^2$, and using the ratio test on this new power series yields the result for the new power series, thus for the original one.
answered Dec 12 '18 at 23:11
MindlackMindlack
2,33217
answered Dec 12 '18 at 23:11
MindlackMindlack
2,33217
answered Dec 12 '18 at 23:11
MindlackMindlack
2,33217
answered Dec 12 '18 at 23:11
MindlackMindlack
2,33217
2,33217
Thanks! One more question: suppose that the radius of convergence would turn out to be finite using the root-test and after rewritting as powers of $x^2$, do I have to take this into account (i.e. does the radius of convergence needs to be squared or something in that case?)
– Student
Dec 12 '18 at 23:27
Yes: if the radius of $sum_n{a_nx^n}$ is $R geq 0$, the radius of $sum_n{a_nx^{2n}}$ is $R^{1/2}$.
– Mindlack
Dec 12 '18 at 23:29
Thanks a lot!..
– Student
Dec 12 '18 at 23:29
add a comment |
Thanks! One more question: suppose that the radius of convergence would turn out to be finite using the root-test and after rewritting as powers of $x^2$, do I have to take this into account (i.e. does the radius of convergence needs to be squared or something in that case?)
– Student
Dec 12 '18 at 23:27
Yes: if the radius of $sum_n{a_nx^n}$ is $R geq 0$, the radius of $sum_n{a_nx^{2n}}$ is $R^{1/2}$.
– Mindlack
Dec 12 '18 at 23:29
Thanks a lot!..
– Student
Dec 12 '18 at 23:29
Thanks! One more question: suppose that the radius of convergence would turn out to be finite using the root-test and after rewritting as powers of $x^2$, do I have to take this into account (i.e. does the radius of convergence needs to be squared or something in that case?)
– Student
Dec 12 '18 at 23:27
Thanks! One more question: suppose that the radius of convergence would turn out to be finite using the root-test and after rewritting as powers of $x^2$, do I have to take this into account (i.e. does the radius of convergence needs to be squared or something in that case?)
– Student
Dec 12 '18 at 23:27
Yes: if the radius of $sum_n{a_nx^n}$ is $R geq 0$, the radius of $sum_n{a_nx^{2n}}$ is $R^{1/2}$.
– Mindlack
Dec 12 '18 at 23:29
Yes: if the radius of $sum_n{a_nx^n}$ is $R geq 0$, the radius of $sum_n{a_nx^{2n}}$ is $R^{1/2}$.
– Mindlack
Dec 12 '18 at 23:29
Thanks a lot!..
– Student
Dec 12 '18 at 23:29
Thanks a lot!..
– Student
Dec 12 '18 at 23:29
add a comment |
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